Question: Simplify; express your answer in exponential form. Assume $a\neq 0, y\neq 0$. $\dfrac{{(a^{-3})^{5}}}{{(a^{3}y^{-1})^{-3}}}$
To start, try working on the numerator and the denominator independently. In the numerator, we have ${a^{-3}}$ to the exponent ${5}$ . Now ${-3 \times 5 = -15}$ , so ${(a^{-3})^{5} = a^{-15}}$ In the denominator, we can use the distributive property of exponents. ${(a^{3}y^{-1})^{-3} = (a^{3})^{-3}(y^{-1})^{-3}}$ Simplify using the same method from the numerator and put the entire equation together. $\dfrac{{(a^{-3})^{5}}}{{(a^{3}y^{-1})^{-3}}} = \dfrac{{a^{-15}}}{{a^{-9}y^{3}}}$ Break up the equation by variable and simplify. $\dfrac{{a^{-15}}}{{a^{-9}y^{3}}} = \dfrac{{a^{-15}}}{{a^{-9}}} \cdot \dfrac{{1}}{{y^{3}}} = a^{{-15} - {(-9)}} \cdot y^{- {3}} = a^{-6}y^{-3}$.